Work Energy And Power Question 43

Question: If the K.E. of a body is increased by 300%, its momentum will increase by [JIPMER 1978; AFMC 1993; RPET 1999; CBSE PMT 2002]

Options:

A) 100%

B) 150%

C) $ \sqrt{300}% $

D) 175%

Show Answer

Answer:

Correct Answer: A

Solution:

Let initial kinetic energy, $ E_1=E $

Final kinetic energy, $ E_2=E+300% $ of E = 4E As $ P\propto \sqrt{E} $

therefore $ \frac{P_2}{P_1}=\sqrt{\frac{E_2}{E_1}}=\sqrt{\frac{4E}{E}}=2 $

therefore $ P_2=2P_1 $

therefore $ P_2=P_1+100% $ of $ P_1 $

i.e. Momentum will increase by 100%.



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